## The Isoperimetric Problem

Problem. Use Lagrange multipliers to determine the function $y=y(x) \in C^{1}[0,1]$, $y(0)=y(1)=0$ that maximizes the area under it and has fixed arc length.

Solution. The arc length $\int_0^1 \sqrt{1+y'^2}dx=\ell$ is given. The functional that we want to minimize has the form $H(y) = \int_0^1 ydx - \lambda \int_0^1 \sqrt{1+y'^2}dx$, Let $h(y,y',\lambda) = y - \lambda \sqrt{1+y'^2}$. There are two ways to do this problem. The first starts with the Euler-Lagrange equations: $\frac{\partial h}{\partial y} - \frac{d}{dx}\big( \frac{\partial h}{\partial y'}\big) = 0$ This directly yields $1 - \lambda \frac{d}{dx}\big(\frac{y'}{\sqrt{1+y'^2}}\big) =0.$ Integrating this yields $x-c= \lambda \frac{y'}{\sqrt{1+y'^2}}$, where $c$ is a constant of integration. Solving this yields $y' = \pm \frac{(x-c)/\lambda}{\sqrt{1-\big((x-c)/\lambda\big)^2}}$ We want to show that $0 < c <1$. Since $y(0)=y(1)=0$ and $y'$ is continuous on $[0,1]$, the mean value theorem implies that there is a point $0<\xi <1$ for which $y'(\xi)=0$. Because $y'$ only vanishes at $c$, it follows that $\xi=c$ and so $0 < c < 1$. In addition, the obvious condition that $y(x) > 0$ on $0 < x < 1$ and the fact that $y(0)=0$ imply that $y'(x) > 0$ for $0 < x < c$, and $y'(x) < 0$ for $c < x < 1$. From this we also see that $\pm \lambda^{-1}$ has to be negative, and so, choosing "$-$" and taking $\lambda>0$, we have $y' = \frac{(c-x)/\lambda}{\sqrt{1-(x-c)^2/\lambda^2}}.$ Doing the integral then gives us $y(x) - \underbrace{y(0)}_{0} = y(x)=\lambda \bigg(\sqrt{1-(x-c)^2/\lambda^2} - \sqrt{1- c^2/\lambda^2}\bigg).$ To determine $c$, set $y(1)=0$ and note that this implies $\sqrt{1-(1-c)^2/\lambda^2} = \sqrt{1-c^2/\lambda^2}$. The only solution $c$ in the interval $0 < c < 1$ is $c=1/2$. With some algebra, we see that $y(x) = \sqrt{\lambda^2-\big(x-\frac12\big)^2} - \sqrt{\lambda^2 - \frac14},$ and so $\big(y + \sqrt{\lambda^2 -\frac14}\big)^2 + \big(x-\frac12\big)^2=\lambda^2$, which is a circle of radius $\lambda$ and center $(\frac12, - \sqrt{\lambda^2 -\frac14})$. Using a little trigonometry, we can show that the angle of the arc or the circle that is cut by rays from the center to $(0,1)$ and to $(1,0)$ is $2\arcsin(\frac{1}{2\lambda})$. The arc length is then $\ell = 2\lambda \arcsin(\frac{1}{2\lambda})$. This equation then uniquely determines $\lambda$; however, it must be solved numerically. One more thing: $\max_{\lambda} 2\lambda \arcsin(\frac{1}{2\lambda}) = \pi$, and so $\ell \le \pi$. However, it's clear that we can use any $\ell$, no matter how large. What's going on?

The second way to do the problem is to note that $h(y,y',\lambda)$ is independent of $x$ and thus has a first integral, $h - y' \frac{\partial h}{\partial y'} = C$. Putting in the explicit form of $h$ and doing some algebra, we obtain $y - \frac{\lambda}{\sqrt{1+y'2}} = C$, from which it follows that $\frac{dx}{dy} = 1/y' = \pm \frac{(y-C)/\lambda}{\sqrt{1 - (y-C)^2/\lambda^2}}$. The rest of the problem is solved in the same way as with the first method.

Updated 1/27/2016 (fjn)